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# If m and n are nonzero integers, is m^n an integer?

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Math Expert
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If m and n are nonzero integers, is m^n an integer?  [#permalink]

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06 Feb 2014, 00:37
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The Official Guide For GMAT® Quantitative Review, 2ND Edition

If m and n are nonzero integers, is m^n an integer?

(1) n^m is positive.
(2) n^m is an integer.

Data Sufficiency
Question: 79
Category: Arithmetic Properties of numbers
Page: 158
Difficulty: 600

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Re: If m and n are nonzero integers, is m^n an integer?  [#permalink]

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06 Feb 2014, 00:38
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SOLUTION

If m and n are nonzero integers, is m^n an integer?

If n is a positive integer then m^n will be an integer for any value of m (taking into account that both are nonzero integers).
If n is negative then m^n will be an integer if and only m=1 or m=-1, for example: (-1)^(-2)=1/(-1)^2=1

So basically we are asked: is n positive or m=|1|?

(1) n^m is positive --> either m=even (and in this case n can take any value) or n=positive (and in this case m can take any value). Not sufficient.

(2) n^m is an integer --> either m=positive (and in this case n can take any value) or m=negative and in this case n=1 or -1. Not sufficient.

(1)+(2) If n^m=(-1)^2=positive integer, then the answer will be NO as m^n=2^(-1)=1/2 but if n^m=1^2=positive integer, then the answer will be YES as m^n=2^1=2. Not Sufficient.

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Re: If m and n are nonzero integers, is m^n an integer?  [#permalink]

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07 Feb 2014, 11:20
1
If m and n are nonzero integers, is m^n an integer?

(1) n^m is positive -> n^m = 2^1 => m^n is an integer; n^m = 2^0.5 => m^n is not an integer
(2) n^m is an integer -> n^m = 2^1 => m^n is an integer; n^m = (-2)^2 => m^n is not an integer.

Combining, n^m = 2^1 => m^n is an integer; n^m = (-2)^2 => m^n is not an integer.

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If m and n are nonzero integers, is m^n an integer?  [#permalink]

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03 May 2015, 12:13
If we want M^n to be an integer n>0 -> If n is negative we'll get a fraction in all cases.

1) N^m > 0 --> 2 scenarios: a) n<0 , m=2 --> n^m >0 b) n>0, m=don't matter - this expression will always be positive, HENCE not sufficient
2) n^m is an integer --> this means that m>0, otherway we'll get a fraction here. We can repeat the steps from 1) --> so, not sufficient.

Update
Variant 2

Let's pick just some smart numbers...
(1) n^m is positive
-$$1^2=2$$, 2^(-1)=1/2 No, $$2^3=8, 3^2=9$$ Yes --> Not Sufficient

(2) n^m is an integer.
Use same examples as abobe. Not Sufficient

1 +2)Using same examples gives us the answer (E). Both statements actually don't give us any new valuable information when combined
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Re: If m and n are nonzero integers, is m^n an integer?  [#permalink]

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03 May 2016, 23:33
Though at first attempt I did it wrong.
But I will try to simplify the solution as much as I can .
is m^n = int ? ..(given m and n are NON ZERO integers)

Now from statement 1 -
n^m is positive.(2 cases - could be an Integer or could a fraction)
case 1 - if m = 2 and n = -2
(-2)^2 = 4 (positive)..now , m^n would be (2)^(-2) = 1/4 (not an Integer.)

case 2 - if m = 1 and n = 2
this will satisfy statement 1 and ofcourse m^n would be 1 (an Integer)
Statement 1 is not sufficient.

From Statement 2 -
n^m is an integer - (It could be an negative integer)
case 1 - if n = -2 and m = 3
m^n would be 3^(-2) which is 1/9 (not an integer)
case 2 - if n = -2 and m = 1
m^n would be 1^(-2) = 1 (an integer).
Statement 2 - not sufficient .

Now statement 1 + 2 says
n^m is an positive integer.
case 1 - if n = -2 and m = 2
m^n would be 1/4 (not an integer)
case 2 - if n = 1 and m = 1
m^n would be 1 (an integer)

Summing up correct option E.
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Re: If m and n are nonzero integers, is m^n an integer?  [#permalink]

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11 Jul 2018, 02:59
Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

If m and n are nonzero integers, is m^n an integer?

(1) n^m is positive.
(2) n^m is an integer.

Given: m, n - integers, m, n are not equal to zero

Statement 1: n^m is positive
n = 1, m = 1, n^m = 1, hence m^n = 1...we get YES
n = -1, m = 2, n^m = 1, hence m^n = 1/2...we get NO

Statement 1 is not Sufficient.

Statement 2: n^m is an integer
n = 1, m = 1, n^m = 1, hence m^n = 1...we get YES
n = -1, m = 2, n^m = 1, hence m^n = 1/2...we get NO

Statement 2 is not Sufficient.

Combining both statements, we can use same examples

Combing is also not Sufficient.

Thanks,
GyM
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Re: If m and n are nonzero integers, is m^n an integer? &nbs [#permalink] 11 Jul 2018, 02:59
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